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How much heat must be removed from 0.15 kg of water to cool it from 95 d/C to 25 d/C? The specific heat capasity of water is 4182 J/kg-k.

mC(T1-T2)
0.15(4182)(70)
= 4.4 x 10^4 J

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asked by Jon
  1. Thinking is correct. I didn't check calc work.

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